hello
I got a Bruno Underground 30 head with two channels. The input jack of channel I is mounted 3 resistors on it(the photo following). It's not like the Low input of a traditional Hi/Low inputs. The 1M resistor seem redundant, is it? The two 68K resistors just form a voltage divider, right?
Thank you.
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I had to stop and think through this at one point but I don't believe any of the resisters are redundant. If you draw both the hi and lo inputs completely with the way they switch when plugged into one or the other, I think it will be more clear.
The 5E3 is the same way. Look at this schematic.
Plugging into the lo input gives you travelling through a 68k resister with a 68K to ground. Plugging into the hi input gives you travelling though a 68K resister with 1Meg to ground.
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bcmatt wrote:I had to stop and think through this at one point but I don't believe any of the resisters are redundant. If you draw both the hi and lo inputs completely with the way they switch when plugged into one or the other, I think it will be more clear.
The 5E3 is the same way. Look at this schematic.
Plugging into the lo input gives you travelling through a 68k resister with a 68K to ground. Plugging into the hi input gives you travelling though a 68K resister with 1Meg to ground.
The amp I got has two channels, each has only one input. The channel I input jack mounted all the 3 resistors. The 5E3 schematic you post is not the same. If the designer wants a Low input, he only needs two 68K resistors to form a voltage divider. Why did he use an additional 1M resistor? Thank you.
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